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finlep [7]
2 years ago
8

The water level in a lake was monitored and was noted to have changed -2 1/3 inches in one year. The next year it was noted to h

ave changed -1 5/6 inches. What was the total change in the water level over the two years?
Mathematics
1 answer:
Diano4ka-milaya [45]2 years ago
5 0

Answer:

The total change in water level over the past 2 years is \displaystyle -4 \frac{1}{6}.

General Formulas and Concepts:

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

Fractions

  • Proper and Improper

Numbers

  • Least Common Multiple (LCM)

Step-by-step explanation:

<u>Step 1: Define</u>

<em>Identify</em>

Year 1 water change:  \displaystyle -2 \frac{1}{3}

Year 2 water change:  \displaystyle -1 \frac{5}{6}

<u>Step 2: Find Total Water Change</u>

  1. [Set up] Add yearly water changes:                                                                \displaystyle -2 \frac{1}{3} + -1 \frac{5}{6}
  2. [Fractions] Convert to improper:                                                                     \displaystyle \frac{-7}{3} - \frac{11}{6}
  3. [Fraction] Rewrite [LCM]:                                                                                  \displaystyle \frac{-14}{6} - \frac{11}{6}
  4. [Order of Operations] Subtract:                                                                       \displaystyle \frac{-25}{6}
  5. [Fraction] Convert to proper:                                                                           \displaystyle -4 \frac{1}{6}
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Finger [1]

Answer:

Consider the following calculations

Step-by-step explanation:

The complete R snippet is as follows

install.packages("Sleuth3")

library("Sleuth3")

attach(case0502)

data(case0502)

## plot

# plots

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main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)

# perform anova analysis

a<- aov(lm(Percent~ Judge,data=case0502))

#summarise the results

summary(a)

### we can use the independent sample t test here

sp<-case0502[which(case0502$Judge=="Spock's"),]

nsp<-case0502[which(case0502$Judge!="Spock's"),]

## perform the test    

t.test(sp$Percent,nsp$Percent)

The results are CHECK THE IMAGE ATTACHED

b)

> summary(a)

Df Sum Sq Mean Sq F value Pr(>F)

Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge

Residuals 39 1864 47.8

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

c)

t.test(sp$Percent,nsp$Percent)

  Welch Two Sample t-test

data: sp$Percent and nsp$Percent

t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

-19.23999 -10.49935

sample estimates:

mean of x mean of y

14.62222 29.49189

6 0
3 years ago
Hi I need help with this homework this is not a test
Lelu [443]

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<em>Note that 7 1/2 is just equal to 7.5.</em>

<em />

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speed=55miles\text{ }per\text{ }hour

Answer:

Ivan was driving at a constant speed of 55 miles per hour.

4 0
1 year ago
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Answer:

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Step-by-step explanation:

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5 0
3 years ago
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Need a bit of help with this question:
DiKsa [7]

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Let use Pythagorean theorem

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\sqrt{581}  = 24.1039... \\  \sqrt{581}  \simeq \: 24.1

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